Question

The work function for iron is 4.50 ev. (a) Find the cutoff wavelength for iron. nm (b) What is the lowest frequency of light incident on iron that releases photoelectrons from its surface? Hz (c) If photons of energy 5.94 eV are incident on iron, what is the maximum kinetic energy of the ejected photoelectrons? ev Need Help? Read It

Answer 1

Given Things

W₀ = 4.50 eV

E = 5.94 eV

h = 4.136 x 10^-15 eV

A)

The cutoff wavelength can be given by

W₀ = hf₀  

where f₀ is the frequency

if i consider the light is moving through air

$W_{0} = \frac{hc}{\lambda }$

$\therefore \lambda = \frac{hc}{W_{0} }$

$\therefore \lambda = \frac{4.136 \times 10^{-15} \times 3\times 10^{8}}{4.50 }$

$\lambda = 275.7 nm$

B)

the frequency and wavelength are related as

$f = \frac{c}{\lambda}= \frac{3 \times 10^{8}}{275.7 \times 10^{-9}}$

$f = 1.088 \times 10^{15} Hz$

C)

By the equation of photoelectric effect

$KE = E - W_{0}$

$KE = 5.94 - 4.50$

$KE = 1.44 eV$